Statements, implication, converse and equivalence | Year 11

01 / LEARN

Understand the idea.

Read the direction of the claim

An implication promises a conclusion whenever its assumption holds. In $P \Rightarrow Q$ , begin by assuming $P$ , then justify $Q$ . Specify the domain first: an integer statement and a real-number statement may have different truth values.

To prove a claim about every multiple of four, represent an arbitrary one as $4 k$ , where $k \in ℤ$ . The letter covers all permitted cases; checking selected numbers only illustrates the claim.

Reverse carefully

The converse swaps the assumption and conclusion. Proving $P \Rightarrow Q$ does not establish $Q \Rightarrow P$ : an even integer need not be divisible by four. A counterexample must satisfy the new assumption but fail the new conclusion.

An equivalence, $P \Leftrightarrow Q$ , requires both directions. Here $P$ is sufficient for $Q$ because it guarantees $Q$ , and necessary for $Q$ because $Q$ cannot hold without $P$ . Write a separate argument for each direction.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

A true implication and a false converse

For an integer $n$ , consider: if $n$ is divisible by 4, then $n$ is even. Prove it and test its converse.

Worked solution

1
Represent an arbitrary multiple of four.
$n = 4 k, k \in ℤ$
Divisibility by four means being four times an integer.
2
Put the number in even form.
$n = 2 (2 k)$
Since $2 k$ is an integer, this is twice an integer; the implication is true.
3
Test the reversed claim.
$6 = 2 \times 3, \frac{6}{4} = \frac{3}{2} \notin ℤ$
Six is even but is not divisible by four, so the converse is false.

Answer

The implication is true; the converse is false.

EXAMPLE 02

An equivalence

For an integer $n$ , prove that $n$ is odd if and only if $n + 1$ is even.

Worked solution

1
Prove the forward direction.
$\begin{aligned} n & = 2 k + 1, k \in ℤ \\ n + 1 & = 2 (k + 1) \end{aligned}$
An odd integer is one more than twice an integer; $k + 1$ is an integer, so $n + 1$ is even.
2
Prove the reverse direction.
$\begin{aligned} n + 1 & = 2 m, m \in ℤ \\ n & = 2 (m - 1) + 1 \end{aligned}$
The integer $m - 1$ shows that $n$ has the required odd form.
3
Combine the two implications.
$n odd \Leftrightarrow n + 1 even$
Both directions have now been established for every integer in the stated domain.

Answer

$n$ odd $\Leftrightarrow n + 1$ even.

03 / YOUR TURN

Try it yourself.

Check your understanding

State the converse of: if $x > 3$ , then $x > 1$ .

Show answer & reasoning

If $x > 1$ , then $x > 3$ .

Swapping assumption and conclusion gives: if $x > 1$ , then $x > 3$ . This converse is false because $x = 2$ satisfies its assumption but not its conclusion.

Keep practising

6 worksheet questions · 14 marks

Open worksheet Worked solutions

Read the questions online

State the assumption and conclusion in: if an integer is divisible by 10, it is divisible by 5.
2 marks
Prove that every multiple of 6 is a multiple of 3.
2 marks
Write the converse of the claim in Question 2 and give a counterexample.
2 marks
For real $x$ , test both directions of $x = 2 \Rightarrow x^{2} = 4$ .
2 marks
For real $x$ , prove $x = 0 \Leftrightarrow x^{2} = 0$ .
3 marks
Is “ $n$ is a multiple of 12” sufficient, necessary, both or neither for “ $n$ is a multiple of 4”, on the integers?
3 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

State the assumption and conclusion in: if an integer is divisible by 10, it is divisible by 5.
2 marks
The condition after “if” is assumed. The statement after “then” must be established.
Assumption: divisible by 10. Conclusion: divisible by 5.
Prove that every multiple of 6 is a multiple of 3.
2 marks
Write $n = 6 k = 3 (2 k)$ for $k \in ℤ$ . Since $2 k$ is an integer, $3 ∣ n$ .
True.
Write the converse of the claim in Question 2 and give a counterexample.
2 marks
3 is divisible by 3 but not by 6.
Every multiple of 3 is a multiple of 6; false at 3.
For real $x$ , test both directions of $x = 2 \Rightarrow x^{2} = 4$ .
2 marks
Substitution gives $2^{2} = 4$ .
For the converse, $x = - 2$ has $x^{2} = 4$ but $x \neq 2$ .
The original is true; the converse is false.
For real $x$ , prove $x = 0 \Leftrightarrow x^{2} = 0$ .
3 marks
If $x = 0$ , squaring gives 0.
If $x^{2} = 0$ , then $x \cdot x = 0$ , so the zero-product property gives $x = 0$ .
Both directions hold.
Is “ $n$ is a multiple of 12” sufficient, necessary, both or neither for “ $n$ is a multiple of 4”, on the integers?
3 marks
Every $12 k = 4 (3 k)$ is divisible by 4.
The number 4 is divisible by 4 but not by 12.
Sufficient but not necessary.

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 1 Mathematics: geometric reasoning and real-number/proof material, regrouped into a school chapter; all exercises are original.

Official curriculum source ↗

In this lesson

Understand the idea.

Read the direction of the claim

Reverse carefully

See the method in action.

A true implication and a false converse

Represent an arbitrary multiple of four.

Put the number in even form.

Test the reversed claim.

An equivalence

Prove the forward direction.

Prove the reverse direction.

Combine the two implications.

Try it yourself.

Check your understanding

Keep practising